Short-Time Fourier Transform (STFT) in Signal Processing - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the function used for STFT from the scipy library.

Signal Processing

from scipy.signal import [1]

Drag options to blanks, or click blank then click option'

Astft

Bspectrogram

Cifft

Dfft

Attempts:

3 left

💡 Hint

Common Mistakes

Importing 'fft' instead of 'stft'.

Using 'spectrogram' which is related but different.

The stft function from scipy.signal computes the Short-Time Fourier Transform.

2fill in blank

medium

Complete the code to compute the STFT of signal 'x' with a window size of 256 samples.

Signal Processing

f, t, Zxx = stft(x, nperseg=[1])

Drag options to blanks, or click blank then click option'

A256

B512

C128

D1024

Attempts:

3 left

💡 Hint

Common Mistakes

Using too small or too large window size causing poor resolution.

Confusing nperseg with other parameters.

The parameter nperseg sets the window size for STFT. Here, 256 samples is used.

3fill in blank

hard

Fix the error in the code to correctly compute the STFT with a sampling frequency of 1000 Hz.

Signal Processing

f, t, Zxx = stft(x, fs=[1])

Drag options to blanks, or click blank then click option'

A50

B1000

C500

D100

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect sampling frequency causing wrong frequency axis.

Confusing fs with window size.

The fs parameter specifies the sampling frequency. Here, it should be 1000 Hz.

4fill in blank

hard

Fill both blanks to compute the STFT with a Hann window and 50% overlap.

Signal Processing

f, t, Zxx = stft(x, window=[1], noverlap=[2])

Drag options to blanks, or click blank then click option'

A'hann'

B128

C256

D'hamming'

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'hamming' instead of 'hann' window.

Setting overlap equal to window size or zero.

The Hann window is specified by 'hann'. For 50% overlap with window size 256, noverlap is 128.

5fill in blank

hard

Fill all three blanks to create a dictionary with frequencies as keys (rounded) and maximum magnitude of STFT for each frequency.

Signal Processing

max_magnitude = [1]([2][i].round().astype(int): abs(Zxx[i]).max() for i, [3] in enumerate(f))

Drag options to blanks, or click blank then click option'

Adict

Bf

Cfreq

Dlist

Attempts:

3 left

💡 Hint

Common Mistakes

Using list instead of dict.

Using wrong variable names causing errors.

The code creates a dictionary (dict) mapping rounded frequencies (f) to max magnitude. The loop variable for frequency is freq.

Practice

1. What is the main purpose of the Short-Time Fourier Transform (STFT) in signal processing?

easy

A. To compress a signal into fewer data points

B. To analyze how frequencies change over time in a signal

C. To remove noise from a signal completely

D. To convert a signal from digital to analog form

Solution

1. Step 1: Understand STFT function

   STFT breaks a signal into small time segments and finds frequency content in each segment.

2. Step 2: Identify main goal

   This helps us see how frequencies change over time, unlike a normal Fourier transform which shows overall frequencies.

3. Final Answer:

   To analyze how frequencies change over time in a signal -> Option B

4. Quick Check:

   STFT = frequency changes over time [OK]

Hint: STFT = frequency analysis in short time windows [OK]

Common Mistakes:

• Confusing STFT with noise removal
• Thinking STFT compresses data
• Mixing STFT with analog conversion

2. Which of the following is the correct Python function call to compute STFT using the scipy library on a signal array named signal with a window size of 256?

easy

A. scipy.stft(signal, nperseg=256)

B. scipy.stft(signal, window=256)

C. scipy.signal.fft(signal, window=256)

D. scipy.signal.stft(signal, nperseg=256)

Solution

1. Step 1: Recall scipy STFT syntax

   The STFT function is in scipy.signal as stft(), and the window size is set by nperseg parameter.

2. Step 2: Match correct syntax

   scipy.signal.stft(signal, nperseg=256) uses scipy.signal.stft(signal, nperseg=256), which is correct syntax.

3. Final Answer:

   scipy.signal.stft(signal, nperseg=256) -> Option D

4. Quick Check:

   Correct function and parameter = scipy.signal.stft(signal, nperseg=256) [OK]

Hint: Use scipy.signal.stft with nperseg for window size [OK]

Common Mistakes:

• Using scipy.stft instead of scipy.signal.stft
• Confusing fft() with stft()
• Passing window size as 'window' instead of 'nperseg'

3. Given the following Python code using scipy.signal.stft, what is the shape of the output array Zxx?

import numpy as np
from scipy.signal import stft
signal = np.sin(2 * np.pi * 5 * np.linspace(0, 1, 100))
f, t, Zxx = stft(signal, nperseg=20)
print(Zxx.shape)

medium

A. (11, 6)

B. (20, 5)

C. (50, 2)

D. (10, 10)

Solution

1. Step 1: Calculate frequency bins

   With nperseg=20, frequency bins = nperseg//2 + 1 = 10 + 1 = 11 rows.

2. Step 2: Calculate time segments

   Signal length=100, window=20, default no overlap means about 6 segments (100/20 rounded up).

3. Final Answer:

   (11, 6) -> Option A

4. Quick Check:

   Zxx shape = (freq bins, time segments) = (11, 6) [OK]

Hint: Output shape = (nperseg/2+1, number of windows) [OK]

Common Mistakes:

• Confusing nperseg with output shape rows
• Miscounting time segments
• Assuming output shape equals input length

4. Identify the error in this STFT code snippet and choose the correct fix:

from scipy.signal import stft
signal = [1, 2, 3, 4, 5]
f, t, Zxx = stft(signal, nperseg=10)
print(Zxx)

medium

A. Error: nperseg is larger than signal length; fix by setting nperseg=5

B. Error: signal must be a numpy array; fix by converting signal to np.array(signal)

C. Error: stft requires window parameter; fix by adding window='hann'

D. No error; code runs fine

Solution

1. Step 1: Check nperseg vs signal length

   Signal length is 5, but nperseg=10 is larger, which causes a warning or error in scipy.

2. Step 2: Fix nperseg value

   Set nperseg to 5 or less to match signal length and avoid error.

3. Final Answer:

   Error: nperseg is larger than signal length; fix by setting nperseg=5 -> Option A

4. Quick Check:

   nperseg ≤ signal length to avoid error [OK]

Hint: nperseg must not exceed signal length [OK]

Common Mistakes:

• Ignoring nperseg size error
• Thinking signal type causes error
• Adding unnecessary window parameter

5. You want to analyze a music clip with fast-changing notes using STFT. Which combination of window size and overlap will give you better time resolution but less frequency resolution?

hard

A. Large window size with high overlap

B. Large window size with low overlap

C. Small window size with high overlap

D. Small window size with low overlap

Solution

1. Step 1: Understand window size effect

   Small window size improves time resolution but reduces frequency resolution.

2. Step 2: Understand overlap effect

   High overlap keeps time segments close, improving time detail further.

3. Final Answer:

   Small window size with high overlap -> Option C

4. Quick Check:

   Better time resolution = small window + high overlap [OK]

Hint: Small window + high overlap = better time detail [OK]

Common Mistakes:

• Choosing large window for better time resolution
• Ignoring overlap effect
• Assuming low overlap improves time resolution